Download Computational Complexity of Fractal Sets 1

Computational Complexity of Fractal Sets
Kamo Hiroyasu
Faculty of Science, Nara Women's University
[email protected]
Kawamura Kiko
Department of Mathematics, University of North Texas
[email protected]
Takeuti Izumi
Graduate School of Informatics, Kyoto University
[email protected]
Abstract
In studies on fractal geometry, it is important to determine whether
the classication by means of computational complexity is independent of the classication by means of fractal dimension. In this paper, we show that each self-similar set dened by polynominal time
computable functions is polynominal time computable, if the selfsimilar set satises a polynominal time open set condition. This fact
provides us examples of sets whose computational complexity are
polynomial time computable, and which have non integer Hausdor
dimension. We also construct a set with computational complexity
NP-complete and with an integer Hausdor dimension. These two
examples establish the independence of computational complexity
and Hausdor dimension.
1 Introduction
The aim of this paper is to nd a mathematical tool other than fractal
dimension to estimate the complexity in fractal geometry.
Although many of the traditional studies of fractal geometry have been made
by means of Hausdor or other dimensions [1], it is not always easy to obtain
the exact value of the dimension. Many of the studies of fractal gures use
self-similar sets, because they are easy to dene, to draw, and to calculate.
The box-counting method often appears in the manipulation of self-similar
sets. Consider a subset A in the unit square [0; 1]2 . We partition [0; 1]2 into
2n 2 2n small squares called pixels and paint each pixel Pi;j obeying the rule
that Pi;j is black if A \ Pi;j 6= and white otherwise. Then we obtain an
approximation of A. By counting the number of black pixels for a large enough
n, we obtain a sucient approximation of box-counting dimension on A.
1
In this paper, we consider both the eciency of approximation and the time
of computation. Although the former has been investigated, e.g. [9], [7], the
latter has never been studied. This is the motivation for investigating the time
to compute an approximation of box-counting dimension as one of the tools for
estimation of complexity in fractal geometry.
Before investigating the time complexity, we must check the computability
of self-similar sets as the rst step. From the point of view of Pour-El{Richards
style computable analysis [8], it is already known that a self-similar set which
is dened by computable contractions is computable [4], [5].
In this paper, we investigate self-similar sets from the viewpoint of computable analysis and propose computational complexity as one of the tools for
estimating the complexity of self-similar sets, other than Hausdor dimension.
First, we recall computational complexity in analysis. The denitions of
computational complexity in this paper are equivalent to those in [11], [12] and
known to be equivalent to those in [8].
Next, we provide a sucient condition for a recursive self-similar set to be
in the computational complexity class P. We obtain two theorems.
Theorem 1 A self-similar set constructed by a set of polynomial time com-
putable functions is in the computational complexity class NP.
Theorem 2 A self-similar set constructed by a set of polynomial time computable functions is in the computational complexity class P, if the self-similar
set satises the polynomial time open set condition.
By using these theorems, we can estimate the computational complexity of
most of the well-known self-similar sets. For examples, the Koch curve is in the
computational complexity class P, and its Hausdor dimension is log 4= log 3,
which is a computable noninteger real number.
As a related work, Ko [6] constructed some gures with P as its computational complexity and an uncomputable number as its Hausdor dimension. He
wasn't concerned with self-similar sets, and his gures were not self-similar sets.
Lastly, we give the following theorem to show an example of NP-complete
subsets of the Euclidean plane. It is the image of a polynomial time computable
Holder function. In addition, we prove that each Holder polynomial time computable function is in the computational complexity class NP. This example
shows that the classication by means of computational complexity does not
coincide with the one by means of Hausdor dimension.
Theorem 3 There exists a closed set in the plane whose computational complexity is NP-complete, and the Hausdor dimension is 1.
In this paper, we use the notation N for the set of all the natural numbers
f0; 1; 2; :::g, Q for the set of all the rational numbers, and R for the set of all
the real numbers
2
2 Complexity of analytical objects
We discuss the computational complexities of real valued functions. The complexity of such a function is determined by a standard encoding of rational
numbers. We have a polynomial time computable standard encoding. First, we
give the formal denition of polyominal time computability.
Denition 2.1 Let f be a function from Nn to N. The function f is polynominal time computable i there exist a Turing machine M and k 2 N such
that M returns the integer f (x1 ; x2 ; : : : ; xn ) as the output in time O((log(x1 1
x2 1 1 1 xn ))k ), if an n-tuple of natural numbers hx1 ; x2 ; : : : ; xn i is given as the
input.
Denition 2.2 The standard encoding of rational numbers is a one-to-one
mapping Q 7! N, denoted q 7! dqe satisying the following conditions.
1. There exists a polynominal time computable function f1 : N2 ! N such
that for each positive integer d and each integer n =
6 0, the natural number
dn=de = f1 (d; n).
2. There exists a polynominal time computable function f2 : N ! N which
satises the following: for each natural number e, it holds that f2 (e) = 0
i there exists a rational number q such that e = dqe.
3. There exist polynominal time computable functions f3 ; f4 : N ! N such
that for each natural number e =
6 0, if f2 (e) = 0 then f3 (e) =6 0 and
e = df4 (e)=f3 (e)e.
Such a encoding q 7! dqe surely exists [10].
We regard dqe as the standard code of the rational number q. For n 2 N,
the size of n is written as s = size (n) 2 N, and is dened by 2s01 n < 2n or
s = n = 0. For q 2 Q, size (q) = size (dqe).
Also we dene the standard encoding of pairing which is polynominal time
computable.
Denition 2.3 The standard encoding of pairing consists of three polynominal
time computable functions: pair : N 2 N 7! N, left : N 7! N, and right : N 7! N
such that
n = pair(left(n); right(n)), m = left(pair(m;n)) and n = right(pair(m; n))
for all natural numbers m; n.
We abbreviate dhm; nie for the pair(m;n) and call dhm; nie the code of the
pair hm; ni. Similarly we abbreviate dhl; m; nie for pair(l; pair(m;n)), and call
dhl; m; nie the code of the triple hl; m; ni. Recursively, we write dhk; l; m; nie
3
for pair(k; pair(l; pair(m; n))) and call dhk; l; m; nie the code of the quadruple
hk; l; m; ni.
We regard dhdq1 e; dq2 e; :::; dqn eie as the encoding of an n-tuple of rational
numbers hq1 ; q2 ; :::; qn i and write simply dhq1 ; q2 ; :::; qn ie.
In general, we use the notation dxe for denoting the coding of a non-naturalnumber object x. That is, for a set X 6= N, the map x 7! dxe : X ! N is a
one-to-one function. Thus, the code of hm; ni 2 N2 is dhm; nie = pair(m; n) 2 N
and the code of hq; ri 2 Q2 is dhq; rie = pair(dqe; dre) 2 N.
We write PN;N for the set of polynominal time computable functions over
natural numbers.
Denition 2.4 The set PN;Q is the set of functions from N to Q such that f
is in PN;Q i there is g 2 PN;N such that df (n)e = g(n) for all n 2 N.
The set PQ;N is the set of functions from Q to N such that f is in PQ;N i
there is g 2 PN;N such that f (q) = g(dqe) for all q 2 Q.
The set PQ;Q is the set of functions of Q into Q such that a function f is
in PQ;N i there is g 2 PN;N such that df (q)e = g(dqe) for all q 2 Q.
A function f 2 PN;Q or f 2 PQ;N or f 2 PQ;Q is called polynominal time
computable.
Similarly we dene the sets PQ2N;Q , PQ;Q2Q , and so forth. The sets PQ2N;Q
is a set of functions from Q 2 N to Q.
Example 2.5 All of the following functions are in PQ2Q;Q.
1. addition: (q; q0 ) 7! q + q0 : Q 2 Q ! Q
2. subtraction: (q; q0 ) 7! q 0 q0 : Q 2 Q ! Q
3. multiplication: (q; q 0 ) 7! q 1 q 0 : Q 2 Q ! Q
0
6= 0) : Q 2 Q ! Q
4. division: (q; q0 ) 7! 0q =q ((qq =
0)
Denition 2.6 The set PR is a subset of R such that r 2 PR i there is a
function q 2 PN;Q such that jq(k) 0 rj < 1=k for each positive integer k.
A real number r 2 PR is called polynominal time computable.
This denition is equivalent to the condition that a real number r belongs to
PR i there exists a Turing machine which computes its approximation with n
signicant digits in polynominal time of n [10] [11] [12].
Example 2.7 All algebraic numbers, , and e are in PR.
Usually, the complexity of a real valued function is dened by the run time of a
Type-2 Machine [10], or an oracle Turing machine [6]. In this paper we dene
complexity in another way. We dene the complexity of a real function by using
4
an ordinal Turing machine, not a Type-2 machine. However, these denitions
are equivalent, that is, the class of polynominal time computable real functions
by our denition is equal to the class of polynominal time computable real
functions by the traditional denition.
Denition 2.8 We say a function f : R ! R is in PR;R i there exist functions g 2 PQ2 ;Q and h 2 PQ2 2N;Q such that the following hold.
1. For each rational number x and each positive rational number > 0, g(x; )
is a rational number such that jg(x; ) 0 f (x)j < :
2. Let x; l be arbitrary rational numbers and k be an arbitrary natural number.
Let z; z 0 be rational numbers which satisfy z 2 (x 0 l; x + l) and z 0 2 (x 0
l; x + l). Suppose jz 0 z0 j < h(x; l; k). Then we have jf (z 0 ) 0 f (z)j < 1=k.
A function f 2 PR;R is called polynominal time computable.
Condition 1 of Denition 2.8 involves the modulus of approximation. Condition
PR2 ;R2 .
2 involves the modulus of uniform continuity. Similary dene the
p set
2 + y 2 , and
The notation jn(x; y)j stands
for
the
absolute
value,
i.e,
j
(
x;
y
)
j
=
x
o
B ((x; y); ) = (x0 ; y0 ) j(x0 ; y0 ) 0 (x; y)j < .
Denition 2.9 2The set2 PR ;R is a set of functions from R2 to R2 such that
a function f : R ! R is in PR ;R i there exist functions g 2 PQ ;Q and
2
2
2
2
3
2
h 2 PQ3 ;Q such that the following hold.
1. For each x 2 Q2 and each rational number > 0, g(x; ) is in Q2 and
jg(x; ) 0 f (x)j < 1=.
2. Let x be an arbitrary rational point 2 Q2 , and ; be arbitrary positive
rational number. Let z; z0 be rational numbers satisfying z 2 B (x; ) and
z0 2 B (x; ). Suppose jz 0 z0 j < h(x; ; ). Then we have jf (z 0 ) 0 f (z )j < .
The following propositions are easily shown.
Proposition 2.10 The set PR is closed under real functions in PR;R .
Proposition 2.11 The set PR;R is closed under composition.
3 Computational complexity of gures
We dene the complexity of geometrical gures. We regard geometrical gures
as closed subsets of R2 . The notation P (X ) stands for the power set of X , that
is, E 2 P (X ) i E X .
5
Denition 3.1 A problem is a subset of N. Let P be a problem, that is, P N
and n be a natural number. If n 2 P then we say that the answer of P for the
instance n is `yes'. If n 62 P then we say that the answer of P for the instance
n is `no'.
Denition 3.2 The set PP (N) is a subset of P (N) such that a problem P is
in PP (N) i there is f 2 PN;N such that P = fn 2 N j f (n) = 0g. A problem
P 2 PP (R) is called polynominal time computable.
Denition 3.3 The set NPP (N) is a subset of P (N) such that a problem P is
in NPP (N) i there is Q 2 PP (N) and f 2 PN;N such that for each n, n 2 P i
there is a number m f (n) such that dhn; mie 2 P . A problem P 2 NPP (N) is
called nondeterministic polynominal time computable, or NP.
Denition 3.4 For problems P and Q01, the relation P P Q holds i there is
a function f 2 PN;N such that P = f (Q). When P P Q, we say that P is
polynominal time reducible into Q.
Denition 3.5 The set NPhardP (N) is a subset of P (N) such that a problem
P is in NPhardP (N) i for any Q 2 NPP (N) , Q P P . The set NPcompP (N) is
a subset of P (N) such that NPcompP (N) = NPP (N) \ NPhardP (N) . A problem
P 2 NPhardP (N) is called NP-hard. A problem P 2 NPcompP (N) is called
NP-complete.
We characterize a closed set X in a plane by determining for given x 2 R2
whether X intersects B (x; ) for each > 0. This means the following. Fix a
point x 2 R2 and a closed set X R2 . Then, does it holds that X \ B (x; ) =
6 for each > 0 ? If yes, then x 2 X . If no, then x 62 X .
6 in general.
Unfortunately, it is not computable whether X \ B (x; ) =
However, for some X , it may be computable whether X intersects to B (x; ) or
is separated from B (x; =2). Therefore, we dene the computable complexity of
closed sets as follows.
Denition 3.6 For a point x 2 R2 and a set X R2 , the notation d(x; X )
stands for the distance between x and X , which is dened as:
d(x; X ) = xinf
jx0 0 xj (X =6 ); d(x; ) = 1:
2X
0
We write F (R2 ) for the set of all closed sets in R2 .
Denition 3.7 Let P be a problem, that is, P N and let X be a closed set
in F (R2 ). We say P ` X i the following conditions hold.
1. If dhx; y; ie 2 P then X \ B ((x; y); 2) 6= . That is, d((x; y); X ) < 2.
2. If dhx; y; ie 62 P then X \ B ((x; y); ) = . That is, d((x; y); X ) > .
6
We say that P determines X i P ` X .
This denition is equivalent to the denition in [10].
Proposition
3.8 For a problem
P and closed sets X; X 0 F (R2 ), if P ` X
0
0
and P ` X , then X = X .
Denition 3.9 The set PF (R2) is a subset of F (R2) such that a closed set X is
in PF (R2 ) i there exists P 2 PP (N) such that P ` X . A closed set X 2 PF (R2 )
is called polynominal time computable.
Denition 3.10 The set NPF (R2 ) is a subset of F (R2) such that a closed set
X is in NPF (R2 ) i there exists P 2 NPP (N) such that P ` X . A closed set
X 2 NPF (R2 ) is called nondeterministic polynominal time computable, or NP.
Denition 3.11 The set NPhardF (R2 ) is a subset of F (R2) such that a closed
set X is in NPF (R2 ) i for any problem P , if P ` X then P 2 NPhardP (N) . The
set NPcompF (R2 ) is a subset of F (R2 ) such that NPcompF (R2 ) = NPF (R2 ) \
NPhardF (R2 ) . A closed set X 2 NPhardF (R2 ) is called NP-hard. A closed set
X 2 NPcompF (R2 ) is called NP-complete.
Lemma 3.12 For a closed set X 2 F (R2), X 2 NPhardF (R ) i there exists
an NP-hard problem P and a polynominal time computable function f of N into
N which satises the following conditions.
1. For each n 2 f (N), there are rational numbers x; y and a positive rational
2
number such that n = dhx; y; ie.
2. If dhx; y; ie 2 f (P ), then d((x; y); X ) .
3. If dhx; y; ie 2 f (N 0 P ), then d((x; y); X ) 2.
Proof. Suppose that Q is a problem and Q ` X . Then we show P P Q, and
therefore Q 2 NPhardP (N) .
If n 2 P , then there is a triple (x; y; ) such that f (n) = dhx; y; ie and
d((x; y); X ) . By the denition of Q ` X , we have f (n) 2 Q. On the other
hand, if n 62 P , then there is a triple (x; y; ) such that f (n) = dhx; y; ie and
d((x; y); X ) 2. By the denition of Q ` X , we have f (n) 62 Q. Therefore
P = f 01 (Q).
2
Example 3.13 Let l; m; n be integers, not all of which are zero. Then a line
lx + my = n is polynominal time computable. The reason is as follows. For each
rational point (x2 =x1 ; y2 =y1 ), the distance d between the point and the line is
calculated as:
s
x2
y2
d = j l 1 =x1 l+2 +mm1 2 =y1 + n j :
7
Then, it is polynominal time computable to determine whether d or d < ,
since this is equivalent to determine whether d2 2 , and this requires only the
comparison of fractions. We can make the computation in polynominal time of
size (x) + size (y) + size ().
4 Self-similar sets
In this section, we discuss the complexities of self-similar sets. We begin with
some preliminaries on contractions and Hausdor metric.
Denition 4.1 Let (V; d) be a metric space. A function over V is a contraction i there is a real number < 1 such that for any points x; y 2 V ,
d((x); (y)) d(x; y). We also say that is a contraction with an upper
bound of magnication < 1.
Denition 4.2 We dene the Hausdor metric between x 2 V and X V ,
and between sets X V and Y V .
0
0
d(x; X ) = yinf
d(x; y); d(X; Y ) = max xinf
d(x ; Y ); yinf
d(y ; X ) :
2X
2X
2Y
0
0
For the empty set, d(x; ) = 1. If either X or Y is empty, then d(X; Y ) = 1.
When V = R2 , this denition of d(x; X ) is identical to Denition 3.6. Note that
for compact sets X; Y , if d(X; Y ) = 0 then X = Y . For a contraction with an
upper bound of magnication < 1, we have d((X ); (Y )) d(X; Y ).
Denition 4.3 A self-similar set is a compact non-empty subset X V which
satises the following set equation for contractions 1 ; 2 ; : : : ; n .
n
[
X = i (X ):
i=1
We call the sequence of the contractions 1 ; 2 ; : : : ; n a self-similar system. If
a compact non-empty subset X V satises this equation, then we say that X
is a self-similar set dened by the self-similar system 1 ; 2 ; : : : ; n .
The fundamental properties of self-similar sets were studied by Hutchinson,
Hata and others [2], [3]. One of the most important properties is the following
lemma.
Lemma 4.4 Suppose that the metric space V is complete. Then for each se-
quence of contractions f1 ; 2 ; : : : ; n g, there exists exactly one self-similar set
dened by 1 ; 2 ; : : : ; n .
8
Proof.
We provide a proof including their proof for completeness and for our
notations.
First we show the construction of the self-similar set. The proof of uniqueness
follows the construction immediately.
Let be the set f1; 2; : : : ; ng. Then l denotes a set of l-tuples of elements
in , that is, l = fhi1 ; i2 ; : : : ; il ij ik 2 for each k g, and ! denotes a set
of innite sequences of elements in , that is, ! = fhi1 ; i2 ; : : : ; ik ; : : :ij ik 2
for each k g. For hi1 ; i2 ; : : : ; il i 2 l , a function hi1 ;i2 ;:::;i i is dened as
l
hi1 ;i2 ;:::;i i (x) = i1 (i2 (: : : (i (x)) : : :)):
For hi1 ; i2 ; : : :i 2 l , a function hi1 ;i2 ;:::i is dened as
hi1 ;i2 ;:::i (x) = llim
(x):
!1 hi1 ;i2 ;:::;i i
l
l
l
For any hi1 ; i2 ; : : :i 2 ! , the limit in the denition of hi1 ;i2 ;:::i always exists. That is because of the following inequalities. Let L be the maximum of
d(x; 1 (x)); d(x; 2 (x)); : : : ; d(x; n (x)). Then this inequation holds:
d( x; i (x) ) L:
l
By applying i 1 ,
l0
By applying i
l0 2
d( i 1 (x); hi
l01
l0
again,
d( hi
l02
;il01 i (x);
hi
;il i (x) ) L:
l02
;il01 ;il i (x) ) 2
L:
After iteration, nally we get:
d( hi1 ;i2 ;:::;i 1 i (x); hi1 ;i2 ;:::;i i (x) ) l01 L:
Therefore, the sequence hi1 ;i2 ;:::;i i (x) converges. Moreover we get the following
l0
l
l
inequality:
d( hi1 ;i2 ;:::;i i (x); hi1 ;i2 ;:::;i ;i +1 ;:::i (x) ) l L=1 0 :
Note that hi1 ;i2 ;:::i (x) does not depends on x, because
d( hi1 ;i2 ;:::i (x); hi1 ;i2 ;:::i (y) ) 1 d(x; y) = 0:
Now we put X as the closure of fhi1 ;i2 ;:::i (x0 )j hi1 ; i2 ; : : :i 2 ! g for a point
x0 . Actually, the set fhi1 ;i2 ;:::i (x0 )j hi1 ; i2 ; : : :i 2 ! g is compact. Note that
X is dened independently of x0 . It is obvious that the set X dened above
satises the set equation
n
[
X = i (X ):
l
l
i=1
9
l
Moreover, if another closed bounded set X 0 also satises this set equation, then
X = X 0 , because d(X; X 0 ) d(i(X ); i (X 0 )) 1d(X; X 0 ), thus d(X; X 0 ) = 0.
2
We dene sets Xl V for natural numbers l as
Xl = fhi1 ;i2 ;:::;i i (x0 )j hi1 ; i2 ; : : : ; il i 2 l g:
These Xl 's depend on x0 . Each Xl is an approximation of X with the estimated
error. We can easily see that d(X; Xl ) l L=(1 0 ).
l
The following proposition is easily proved.
Proposition 4.5 For a self-similar set X = S i(X ) (1 i n), the Hausdor dimension of X is 0 log n= log .
We have dened contractions and self-similar sets in a general complete
metric space V . Hereafter we discuss only the case where V = R2 , the Euclidean
plane, and contractions which are functions from R2 to R2 . We regard closed
sets as gures on R2 .
The notion of computational complexity does not appear in the denition
of self-similar sets. Now we dene the notion of a self-similar system with
complexity.
Denition 4.6 A self-similar system (1; 2; : : : ; n) is called polynominal
time computable, or a P-self-similar system, i all of i 's are in PR2 ;R2 .
Proposition 4.7 If all of 1 ; 2; : : : ; n 2 PR2 ;R2 are contractions, then each
hi1 ;i2 ;:::;i i is also in PR2 ;R2 . Moreover, an approximation of hi1 ;i2 ;:::;i i (x)
within an error less than 1=2m is computable in polynominal time of l + m +
size (x) for each rational point x.
Proof. Let fi be an approximation to i such that ji(x; ) 0 fi (x; )j < for each i. There is an integer k such that fi (x:) is computable in the time
O((size (x) + size ())k ), for all i, because each i is in PQ3 ;Q2 . Moreover we
assume that dn=de is computable in the time O((size (d) + size (n))k ).
We will calculate the approximation of hi1 ;i2 ;:::;i i (x) in several steps with
auxiliary variables y1 ; y2 ; : : : ; yl . We will calculate ys in Step s for 1 s l
l
l
l
recursively.
In this step we will calculate y1 . For any l; m, we calculate d1=l12 e,
which is computable in time O((size (l 1 2m ))k ) O((m 1 log l)k ). Therefore
size (1=l12 ) O((m 1 log l)k ). Then we calculate y1 = fi (x; 1=l12 ). which
is
2
1
k
k
computable in time O((size (x) + size ( =l12 )) ) O((size (x) + l 1 log m) ). It
holds that jy1 0 i j 1=l12 . Note that size (y1 ) O((size (x) + size (l 1 2m ))k ) O((size (x) + l 1 log m)k ), because y1 is as large as x, and the precision of y1 is
1=
l12 .
Step s. (2 s l). In this step, the variables y1; y2; 1 1 1 ; ys01 have been
calculated. Now we calculate ys .
Step 1.
m
m
l
m
l
m
m
10
m
0
1
0
1
size (ys01) O (size (x) + size (l 1 2m ))k O (size (x) + l 1 log m)k ;
jy1 0 hi
l0s+2
Then, we calculate ys = fi
l0s+1
;il0s+3 ;:::;il i (x)j <
s01 :
l 1 2m
(ys01 ; 1=l12 ), which is computable in time
m
O((size (ys01 ) + size (l 1 2m ))k22)
O((size (x) + 2size (l 1 2m ))k ) O((size (x) + l 1 m)k2 ):
Then it holds that
1
l 1 2m :
Note that size (ys ) is also O(size (x) + size (l 1 2m )) O(size (x) + l 1 m), which
is independent to s. That is because jys j is as large as jxj and the precision of
y1 is 1=l12 .
Finally, we obtain yl which is the approximation of ~i(x) with error 1=2m .
2
The total time of calculation is O((m log l) + l 1 (size (x) + m log l)k ) O((l +
m + size(x))2k+1).
2
Each i is a contraction. This fact prohibits the error of each step i from
jys 0 hi
l0s+1
;il0s+2 ;:::;il i j m
k
growing too big.
Theorem 1 A self-similar set dened by a P-self-similar system is in NPF (R ).
Proof. For the given self-similar set X = S i (X ) and a rational point x0, we
2
construct the following NP problem.
For a given rational point x = (x2 =x1 ; y2 =y1 ) and a given small
rational number = e2 =e1 > 0, determine whether there exists l
and ~i 2 l satisfying ~i(x) 0 x < 3=2, where ~i (x) is the rational
point which is an approximation of ~i(x) within an error < =4,
and l is an integer such that 2 =4 < l L=(1 0 ) < =4 for L =
maxf ji (x0 ) 0 x0 j gi .
The integer l can be calculated in the following manner: any integer l such that
log((1 0 )=4L) 0 2 < l < log((1 0 )=4L)
0 log 0 log will work. This is computable in polynominal time of size() O((size(e1 ) +
size (e2 ))k , because and L are constant numbers.
11
First, we note that this problem is indeed an NP problem, that is, this
question is computable in the polynominal time of size (x) + size (). Then we
will show the problem determines the self-similar set X .
Put z = ~i (x). If the ~i above exists, then j~i(x0 ) 0 xj j~i(x0 ) 0 z j +
jz 0 xj < 7=4. Hence the Hausdor distance d(x; Xl ) is less than 7=4. And
d(X; Xl ) < l L=(1 0 ) < =4. Therefore d(x; X ) < 2.
To the contrary, suppose that there does not exist such ~i. Then for each
~i, the distance is estimated as j~i(x0 ) 0 xj jz 0 xj 0 j~i(x0 ) 0 z j > 5=4.
Thus d(x; Xl ) > 5=4. Hence the Hausdor distance is estimated as d(x; X ) d(x; Xl ) 0 d(X; Xl ) > .
Thus this problem determines X .
2
Next, we give more precise estimations of computational complexity for selfsimilar sets with some condition. Instead of Theorem 1, we know many polynominal time computable self-similar sets dened by P-self-similar systems. One
of the most famous of them is the Koch curve. We will give a sucient condition
for such self-similar sets.
Denition 4.8 Let (1; S2; : : : ; n) be a self-similar system, and X be a self-
similar set such that X = i(X ). Then the self-similar system (1 ; 2 ; : : : ; n )
satises the open set condition i there is an open set W such that
X W and i (W ) \ j (W ) = for i 6= j:
The Koch curve is dened by a self-similar system which satises the open
set condition. Yet, this open set condition is not sucient to determine the
complexity of self-similar sets, because the open set condition does not have the
notion of complexity. We have to give a stronger denition.
Denition 4.9 Let (1; S2; : : : ; n) be a self-similar system, and X be a self-
similar set such that X = i (X ). The the self-similar system (1 ; 2 ; : : : ; n )
satises the polynominal time open set condition, or the P-open-set condition, i there exists an open set W , a real number 2 PR and functions
k 2 PQ3 2N2 ;N and j 2 PQ3 2N4 ;N which satisfy the following:
1. X W .
2. i (W ) \ j (W ) = for i =
6 j.
3. ji (y) 0 i (x)j jy 0 xj for each i 2 and x; y 2 R2 .
4. For each rational point x = (x2 =x1 ; y2 =y1 ), each positive natural number
c > 0, a natural number l, and a sequence ~j 2 l , B (x; l c) \ ~j (W ) =
6 ,
l
there is an integer i such that 0 i < k(x; c; 2 ) and
~j = j (x; 2c ; 2l ; i; 1); j (x; 2c; 2l ; i; 2); : : : ; j (x; 2c ; 2l ; i; l) :
12
The last condition says that we can list all the ~i 2 l such that B (x; ) \ ~i(W ) 6=
, that is, d(x; ~i (W )) < , in polynominal time of size (x) + 1= + l.
Theorem 2 A P-self-similar set which satises the P-open-set condition is in
PF (R2 ) .
Proof.
In the NP problem in the proof of Theorem 1, the possible choices of
~i 2 l are listed in polynominal time if we put c as a constant 4L=2 (10) .
Therefore, we can check on all the possible choices in polynominal time.
2
Most of the self-similar sets which have been well analysed satisfy the P-open-set
condition, hence they are in P. Then the following question arises.
Question: Is there a P-self-similar set in NPF (R2 ) 0 PF (R2) , when we
assume P =
6 NP ?
The answer has not been obtained yet; however, in the next section, we will
construct an NP complete set, although it is not self-similar.
As the next lemma shows, Theorem 2 is useful enough, because many well
known self-similar sets satisfy the P-open-set condition.
Lemma 4.10 Let (1; 2; : : : ; n ) be a P-self-similar system consisting similar
maps of common magnication , satisfying the open set condition. Then, it
satises the P-open-set condition.
Proof. First we note that 2 PR, because is the common magnication of
i 's, which are in PR2 ;R2 .
Let X be the self-similar set dened by (i )i , and W the open set which
appears in the statement of open set condition. Let d be a positive rational
number
8 which is greater
9 than or equal to the diameter of W , that is, d sup jy 0 xj x; y 2 X . Let a be a positive rational number which is less
than or equal to the area of W . Let x0 be a rational point in W . This point
x0 plays the role of x0 in the proof of Lemma 4.4. Fix a rational number
such that l c=2 l c, where l c is as in Denition 4.9. Note that
size () O((l + size (c))h ) for some h, because is a constant.
We construct functions j and k which list the possible choices of i. Let K
be an integer such that K n and K (4c + 7d)2 =a. The function k(x; c; m)
is a function whose value is K , which depends only on c and is independent of
x and m. It is obvious that k 2 PQ3 2N;N .
Let fi be a function in PQ3 ;Q2 such that ji (x) 0 fi (x; )j < for each
rational point x and positive rational number . There is 0an integer h such that
h1
for each i, the value fi (x;
0 ) is computable hin1 the time O (size (x) + size ()) ,
and size (fi (x; )) O (size (x) + size ()) , because each i is polynominal
time computable. Now, we show the procedure for calculating j . The procesure
consists of several steps. In Step s for 1 s l, we will recursively calculate an
13
approximation of s , written s , and a mapping js (m; t) which maps (m; t) 2
f1; 2; : : : ; sg 2 f1; 2; : : : ; ksg into js (m; t) 2 f1; 2; : : : ; ng. We will write ~js;t as
~js;t = hjs (1; t); js(2; t); : : : ; js (s; t)i 2 f1; 2; : : : ; ngs
Step 1. First we calculate , which is an approximation
of with error < =2l,
and < 1. The calculation time is O((size (l)h )) for some h, because is
constant and 2 PR .
Next, for each i such that 1 i n, we calculate fi (x0 ; =l). Each fi is an
approximation for i (x0 ). Thus we write i (x0 ) for fi (x0 ; =l). We put k1 = n
and j1 (1; t) = t for 1 t k1 .
Step s. (2 s l). Assume we have calculated s01 ; js01(m; t) and ~j 1 (x0)
for 1 m s 0 1; 1 t ks01 . We write ~js01;t as
~js01;t = hjs01(1; t); js01 (2; t); : : : ; js01 (s 0 1; t)i;
s0 ;t
As induction hypothesis, they satisfy the following:
~j
s01;t
(x0 ) 0 ~j
s01;t
(x0 ) < (s 0 1)=l;
and, for each ~i = hi1 ; i2 ; : : : ; il i 2 l , there exists t such that ~js01;t = hil0s+2 ;
il0s+3; : : : ; il i if jx 0 ~i(x0 )j < .
Now, we calculate s , which is obtained from 2 s01 . Note that js 0 s j <
s=l, because j 0 j < =l. Next we calculate fi ~j 1 (x0 ); =l for each i and
t, where 1 i n and 1 t ks01 . We write hi;~j 1 i (x0 ) for fi ~j 1 (x0 ) .
Then it holds that
s0 ;t
s0 ;t
hi;~j
s01;t
s0 ;t
i (x0 ) 0 hi;~j
s01;t
i (x0 ) < s=l:
Hence, there is an integer ks K such that there are ks pairs of (t; i) such that
hi;~j
s
i (x0 ) 0 x < 3 + d:
hi;~j
s
i (x0 ) 0 x < 3 + 4 d;
s01;t
Since
s01;t
we have
hi;~j
s01;t
s
(
x
)
0
x
< l + 3 + 4s d < 4 + 6s d < s(4c + 6d):
i 0
On the other hand, we have x0 2 W . Thus
hi;~j
s01;t
s
i (W ) B (x; (4c + 7d)):
14
The number of pairs (i; t) which satises the last inequality above is (4c +
7d)2 =a K . We enumerate such i's and t's as i1 ; i2 ; : : : ; ik and t1 ; t2 ; : : : ; tk .
Thus, we enumerate such (t; i)'s as (t1 ; i1 ); (t2 ; i2 ); : : : ; (tk ; ik ). We dene
js (u; m) for 1 m s; 1 u ks by
s
s
s
s
js (u; 1) = iu ; js(u; m) = js01 (tu ; m + 1) (m s 0 1):
Put ~js;u = hiu ; ~jt ;s01 i = hjs (u; 1); js (u; 2); : : : ; js (u; s 0 1)i.
Note that for each ~i = hi01 ; i02 ; : : : ; i0l i 2 l , if ~i (W ) \ B (x; l c) 6= , then
there exists u such that ~js;t = hi0l0s+1 ; i0l0s+2 ; : : : ; i0l i: This is because if ~i(W ) \
B (x; l c) =
6 , then ~i(W ) B (x; l c + l d), and ~i (X ) B (x; l c + l d),
u
u
therefore
hi
0
l
l
s
;i0l0s+2 ;:::;i0l i (W ) B (x; c + d + d):
hi
0
;i0l0s+2 ;:::;i0l i (x0 ) 2 hi0l0s+1 ;i0l0s+2 ;:::;i0l i (W );
l0s+1
We also have
l0s+1
hi
and
0
l0s+1
Therefore
hi
;il0s
0
(x ) 0 hi
+2 ;:::;i i 0
0
l
0
l0s+1
;il0s
0
(x ) < s=l:
+2 ;:::;i i 0
0
l
l
l
s
s
s
;i0
;:::;i0l i (x0 ) 0 x < s=l + c + d + d < 3 + 2 d < 3 + 4 d:
l0s+1 l0s+2
0
Last cstep.
l
In this step, we put j (x; 2c ; 2l ; s; t) = jl (s; t) for 1 t kl , and
j (x; 2 ; 2 ; s; t) = 1 otherwise.
We have calculated ~j (x0 ) for 1 t ks K . The size of the value of
0
1
each ~j (x0 ) is as large as O (size (x0 ) + size ())h , because the precision of it
is near , that is, it has the error as large as O(). Therefore, j is a polynominal
time computable function.
2
Example 4.11 The Koch curve satises our P-open-set condition.
s;t
s;t
5 The image of a function satisfying a Holder
condition
In this section, we construct an NP complete set which is dened as the image
of a function satisfying a Holder condition.
Denition 5.1 Let V and W be metric spaces. A function f from V to W
satises a Holder condition of index i there is a k such that for any x; y 2 V ,
dW ( f (x); f (y) ) < k(dV (x; y)) :
15
Functions satisfying a Holder condition are continuous. A Holder condition of
index 1 is equivalent to Lipschitz continuity.
Proposition 5.2 Let f be a function from V to W satisfying a Holder condition of index , and the Hausdor dimension of V is D. Then the Hausdor
dimension of f (V ) is D=.
The proof is clear.
Hereafter, we write I for a unit interval [0; 1].
Denition 5.3 The set PI;R is the set of functions from I to R such that f is
in PI;R i there is g 2 PR;R and 8x 2 I:f (x) = g(x).
The set PI;R2 is the set of functions from I to R2 such that f is in PI;R2
i there is f1 ; f2 2 PI;R and f (x) = (f1 (x); f2 (x)).
Lemma 5.4 Let f be a function in PI;R2 If f satises a Holder condition of
some index, then f (I ) is in NPF (R2 ) .
Proof. If ~i = hi1; i2; : : : ; il i 2 f0; 1gl is the bit sequence, then we write 0:~i for
the binary fraction 0:i1 i2 : : : il .
Now we construct an NP problem such that:
Let x = (x2 =x1 ; y2 =y1 ) and = e2 =e1 > 0 be given. The question is
whether there exists some choice ~i 2 l which satises jy 0 xj < 3=4
where y is an approximation of f (0:~i) within an error < =4, and
k=2l < =4.
If such an i exists, then the distance jy 0 xj is < 3=4. And d(y; f (I )) < =4.
Therefore d(x; f (I )) < . If not, then jy 0 xj 3=4. Therefore d(x; f (I )) > =2.
Thus this problem determines f (I ).
2
There exists the image of a function satisfying a Holder condition in NP and
not in P. Indeed, it is in the complexity of NP complete. We construct such an
orbit in the remaining part of this paper. Of course we assume P 6= NP.
We take an NP complete problem Q which consists of the following data:
For a given instance ~v 2 f0; 1gl for some l, the question is whether
there exists some guess w~ 2 f0; 1gl which satises q(~v; w~ ) = 0, where
q is polynominal time computable function of (f0; 1g3 )2 into f0; 1g.
In this problem Q, the lengths of an instance v and the certicate w for the
instance v are equal.
Next we construct a function of I into R2 satisfying a Holder condition such
that the problem Q is reduced to f (I ). The function f consists of two functions
f1 and f2 of I into R such that f (t) = (f1 (t); f2 (t)).
After we dene f , we will reduce the NP problem Q into a problem P which
determines f (I ). For this purpose, we dene an encoding c(~v; w~ ) which maps binary sequences ~v 2 hv1 ; v2 ; : : : ; vl i 2 f0; 1gl and w~ 2 hw1 ; w2 ; : : : ; wm i 2 f0; 1gm
16
to a quinary sequence c(~v; w~ ) = hc1 ; c2 ; : : : ; cl+m+1 i 2 f0; 1; : : : ; 4gl+m+1 . The
encoding c(~v; w~ ) is dened in the following manner.
1. ci = vi + 2 for 1 i l.
2. cl+1 = 1.
3. ci = wi0l01 + 2 for l + 2 i l + m + 1.
For example, c(h0i; h1i) = h213i, and c(h10i; h110i) = h321332i. Note that
c(~v; w~ ) = hc1 ; : : : :cl+m+1 i consists of only 1, 2, and 3. It does not have 0 or 4.
We use the notations 0:~v and (0:~c)5 as follows. If ~v is a binary sequence, then
0:~v denotes a binary fraction 0:v1 v2 : : : vl = 201 v1 +202 v2 +1 +20l vl . Similarly, if
~c = hc1 ; c2 ; : : : ; cn i is a quinary sequence, then (0:~c)5 denotes a quinary fraction
(0:c1 c2 : : : cn )5 = 501 c1 + 502 c2 + 1 + 50n cn .
First we dene f1 . Because f is continuous, so is f1 . Therefore, it is sucient
to dene the value of f1 (t) only for quinary fractions t. Let t 2 I be the t =
(0:t1 t2 : : : tn )5 , where tn 6= 0. Let i be a smallest number such that ti 2 f0; 4g,
that is, tj 2 f1; 2; 3g for all j < i. Then we make the case selection by using
this i.
Case 1: If t = 1, that is, there does not exist a sequence ~t such that
t = (0:~t)5 , then:
f1 (1) = 0.
Case 2: If such a digit ti does not exist, that is, all of tj 's are 1, 2 or 3,
then:
{ Case 2.1: If there exist binary sequences ~v and w~ such that q(~v; w~ ) =
0 and ht1 ; : : : ; tn i = c(~v; w~ ), then:
f1 (t) = 1=5n . Note that n 1 in this case.
{ Case 2.2: Otherwise, f1(t) = 0. This case includes t = 0.
Case 3: If such a digit ti exists, that is, i is a smallest number such that
ti is 0 or 4, then:
{ Case 3.1: If ti = 0, then:
f1 (t) = (1 0 u)f1 ( (0:t1 : : : ti01 )5 ), where u = (0:ti+1 : : : tn )5 .
{ Case 3.2: If ti = 4, then:
f1 (t) = u 1 f1 ( (0:t1 : : : ti02(ti01 + 1))5 ), where u = (0:ti+1 : : : tn )5 , for
i 2.
If i = 1,then f (t) = 0.
It is easy to observe that f1 satises the Lipschitz condition, because jf1 (t0 ) 0
f1 (t)j jt0 0 tj: If the pair of ~v and w~ is a valid pair of an instance and a
guess, that is, q(~v; w~ ) = 0 holds, then f1 (t) behaves as the following graph near
t0 = (0:c(~v; w~ ))5 .
17
6f (t)
1
1=5n
0@@
0
@@
00
@@
00n
t0 0 1=5
t0
t0
+ 1=5n
-t
Next we dene f2 , which is similar to f1 .
Case 1: If t = 1, that is, there does not exist a sequence ~t such that
t = (0:~t)5 , then:
f2 (1) = 0.
Case 2: If such a digit ti does not exist, that is, all of tj 's are 1, 2 or 3,
then:
{ Case 2.1: If there exist binary sequences ~v and w~ such that q(~v; w~ ) =
0 and ht1 ; : : : ; tn i = c(~v; w~ ), then:
f2 (t) = (0:~v)=5n . Note that n 1 in this case.
{ Case 2.2: Otherwise, f2(t) = 0. This case includes t = 0.
Case 3: If such a digit ti exists, that is, i is a smallest number such that
ti is 0 or 4, then:
{ Case 3.1: If ti = 0, then:
f2 (t) = (1 0 u)f2 ( (0:t1 : : : ti01 )5 ), where u = (0:ti+1 : : : tn )5 .
{ Case 3.2: If ti = 4, then:
f2 (t) = uf2 ( (0:t1 : : : ti02 (ti01 + 1))5 ), where u = (0:ti+1 : : : tn )5 , for
i 2.
If i = 1,then f (t) = 0.
Let t0 be (0:c(~v; w~ ))5 , and l be the length of ~v;. Here the length of c(~v; w~ ) is
2l + 1. If the pair of ~v and w~ is a valid pair of an instance and a guess, that is,
q(~v; w~ ) = 0 holds, then the part of the orbit f ([t0 0 1=52l+2 ; t0 + 1=52l+2 ]) draws
the following segment.
18
f2 (t) 6
" (1=5 l ; (0:~v)=5 l
"
""
"
""
"
"
"
O
l
(0:~v)=52l+2
2 +2
1=52 +2
2 +2
)
f1 (t)
It is clear that f (I ) is the union of such segments for all ~v's which have their
own certicates.
Theorem 3 There is a function f in PI;R2 , which satises the Lipschitz continuity, and whose image f (I ) is in NPcompF (R2 ) .
Proof. We show the function f dened as above satises the statement of this
theorem.
It is obvious that f 2 PI;R2 . By Lemma 5.4, f (I ) 2 NPF (R2 ) .
Next we will show that f (I ) 2 NPhardF (R2 ) . In order to show that, we need
to nd a function g of f0; 1g3 into R which satises the condition in Lemma 3.12.
We are identifying f0; 1g3 with N by the standard isomorphism f0; 1g3 = N.
Remember that the problem Q is dened as follows. For an instance ~v 2
f0; 1g3 , ~v 2 Q i there is some w~ 2 f0; 1g3 of the same length as ~v such that
q(~v;~;w) = 0.
We dene g as g(~v) = dh1=52l+2 ; (0;~v)=52l+2 ; 1=53l+3 ie, where l is the length
of ~v. It is certain that g is in P. Note that g is an one-to-one map.
If f (~v) 2 f (Q), that is, ~v 2 Q, then there is some w~ such that q(~v; w~ ) = 0.
Let a point p be where l is a length of ~v. Then f (I ) passes through the point
(1=52l+2 ; (0:~v)=52l+2 ). Hence d( (1=52l+2 ; (0:~v)=52l+2 ); f (I ) ) = 0.
On the other hand, if f (~v) 2 f (N 0 Q), that is ~v 2 Q, then there is no w~
such that q(~v; w~ ) = 0. Then f (I ) does not have a segment whose inclination is
0:~v. Therefore, f (I ) does not pass through the point (1=52l+2 ; (0:~v)=52l+2 ). The
nearest segments to this point is the segment through the point (1=52l+2 ; (0:~v 0
1=5l )=52l+2 ) and the segment for point (1=52l+2 ; (0:~v + 1=5l )=52l+2 ). Other
segments are in more dierent inclinations, or too short in the length. Therefore
d( (1=52l+2 ; (0:~v 0 1=5l )=52l+2 ); f (I ) ) > 1=53l+3 .
Thus, by Lemma 3.12, f (I ) is in NPhardF (R2 ) .
2
6 Conclusion
We have shown three major results. First is Theorem 1, which states that the
complexity of each self-similar set dened by a P-self-similar system is nondeterministic polynominal time computable, that is, the gure is in NPF (R2 ) . The
19
second is Theorem 2, which states that the complexity of each self-similar set
dened by a P-self-similar system with the P-open-set condition is polynominal time computable, that is, the gure is in PF (R2 ) . The third is Theorem
3, which provides an example of the image of a polynominal time computable
Holder function such that its Hausdor dimension is 1 and its computational
complexity is NP-complete.
Thus we conclude that the classication by means of computable complexity
does not coincide with the classication by means of the Hausdor dimension.
Acknowledgement
The authors would like to thank Prof. Yamaguchi Masaya, Prof. Yasugi Mariko,
and Prof. Klaus Weihrauch for their comments.
We especially express our gratitude to Prof. Karen Brucks for her extensive
help in preparing this paper.
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